The answer to this is surface wave i think
Answer:
Lose two electrons.
Explanation:
Barium is present in group 2.
It is alkaline earth metal.
Its atomic number is 56.
Its electronic configuration is Ba₅₆ = [Xe] 6s².
In order to attain the noble gas electronic configuration it must loses its two valance electrons.
When barium loses it two electron its electronic configuration will equal to the Xenon.
The atomic number of xenon is 54 so barium must loses two electrons to becomes equal to the xenon.
Answer:
669.48 kJ
Explanation:
According to the question, we are required to determine the heat change involved.
We know that, heat change is given by the formula;
Heat change = Mass × change in temperature × Specific heat
In this case;
Change in temperature = Final temp - initial temp
= 99.7°C - 20°C
= 79.7° C
Mass of water is 2000 g ( 2000 mL × 1 g/mL)
Specific heat of water is 4.2 J/g°C
Therefore;
Heat change = 2000 g × 79.7 °C × 4.2 J/g°C
= 669,480 joules
But, 1 kJ = 1000 J
Therefore, heat change is 669.48 kJ
Answer:
![T_{eq}=28.9\°C](https://tex.z-dn.net/?f=T_%7Beq%7D%3D28.9%5C%C2%B0C)
Explanation:
Hello!
In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:
![Q_{Cd}+Q_{W}=0](https://tex.z-dn.net/?f=Q_%7BCd%7D%2BQ_%7BW%7D%3D0)
Thus, we insert mass, specific heat and temperatures to obtain:
![m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0](https://tex.z-dn.net/?f=m_%7BCd%7DC_%7BCd%7D%28T_%7Beq%7D-T_%7BCd%7D%29%2Bm_%7BW%7DC_%7BW%7D%28T_%7Beq%7D-T_%7BW%7D%29%3D0)
In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:
![T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}](https://tex.z-dn.net/?f=T_%7Beq%7D%3D%5Cfrac%7Bm_%7BCd%7DC_%7BCd%7DT_%7BCd%7D%2Bm_%7BW%7DC_%7BW%7DT_%7BW%7D%7D%7Bm_%7BCd%7DC_%7BCd%7D%2Bm_%7BW%7DC_%7BW%7D%7D)
Now, we plug in to obtain:
![T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C](https://tex.z-dn.net/?f=T_%7Beq%7D%3D%5Cfrac%7B37.60g%2A0.232%5Cfrac%7BJ%7D%7Bg%5C%C2%B0C%7D%2A100.00%5C%C2%B0C%2B25.00g%2A4.184%5Cfrac%7BJ%7D%7Bg%5C%C2%B0C%7D%2A23.0%5C%C2%B0C%7D%7B37.60g%2A0.232%5Cfrac%7BJ%7D%7Bg%5C%C2%B0C%7D%2B25.00g%2A4.184%5Cfrac%7BJ%7D%7Bg%5C%C2%B0C%7D%7D%5C%5C%5C%5CT_%7Beq%7D%3D28.9%5C%C2%B0C)
NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.
Best regards!
Answer:
As the kinetic energy of the gaseous solute increases, its molecules have a greater tendency to escape the attraction of the solvent molecules and return to the gas phase. Therefore, the solubility of a gas decreases as the temperature increases.
Explanation:
As the kinetic energy of the gaseous solute increases, its molecules have a greater tendency to escape the attraction of the solvent molecules and return to the gas phase. Therefore, the solubility of a gas decreases as the temperature increases